A projectile is launched horizontally with a speed of 80.0 m/s. If the projectile is launched 1.5m above the floor, how long does it take the projectile to hit the floor?
To find the time it takes for the projectile to hit the floor, we can use the equation of motion:
\[
y = y_0 + v_{0y}t + \frac{1}{2}gt^2
\]
Where:
- \(y\) is the final vertical position (in this case, 0m, since the floor is at y=0)
- \(y_0\) is the initial vertical position (1.5m above the floor)
- \(v_{0y}\) is the initial vertical velocity (in this case, 0 m/s since the projectile is launched horizontally)
- \(g\) is the acceleration due to gravity (approximately 9.8 m/s\(^2\))
- \(t\) is the time it takes for the projectile to hit the floor
Since the projectile is launched horizontally, the initial vertical velocity is 0. Thus, the equation simplifies to:
\[
0 = 1.5 + \frac{1}{2}(9.8)t^2
\]
To find \(t\), we can rearrange the equation and solve for \(t\):
\[
\frac{1}{2}(9.8)t^2 = -1.5
\]
Divide both sides by \(\frac{1}{2}(9.8)\):
\[
t^2 = \frac{-1.5}{\frac{1}{2}(9.8)}
\]
Simplify the right side:
\[
t^2 = \frac{-1.5}{4.9}
\]
Take the square root of both sides to isolate \(t\):
\[
t = \sqrt{\frac{-1.5}{4.9}}
\]
Calculating this, we get:
\[
t \approx 0.39 \, \text{s}
\]
Therefore, it takes approximately 0.39 seconds for the projectile to hit the floor.
The horizontal launching speed has zero vertical component. The time to hit the ground will be the same as if it were simply dropped.
1.5 m = (1/2) g t^2
Solve for t